The enthalpy of a reaction at 273 K. is -3.57 KJ. What will be the enthalpy of reaction at 373 K if `DeltaC_(p)` = zero :-

Options

A. `-3.57`
B. Zero
C. `-3.57xx(373)/(273)`
D. `-375`

Answer

To solve the problem of finding the enthalpy of a reaction at a different temperature given the enthalpy at a known temperature and the change in heat capacity (ΔCp), we can use the Kirchhoff equation. Here’s a step-by-step solution:

### Step 1: Identify the Given Values
- **Initial Temperature (T1)** = 273 K
- **Final Temperature (T2)** = 373 K
- **Enthalpy at T1 (ΔH1)** = -3.57 kJ
- **Change in Heat Capacity (ΔCp)** = 0

### Step 2: Write the Kirchhoff Equation
The Kirchhoff equation relates the change in enthalpy with temperature and heat capacity:
\[ \Delta H(T_2) = \Delta H(T_1) + \Delta C_p \cdot (T_2 - T_1) \]

### Step 3: Substitute the Known Values
Since ΔCp = 0, the equation simplifies to:
\[ \Delta H(T_2) = \Delta H(T_1) + 0 \cdot (T_2 - T_1) \]
This means:
\[ \Delta H(T_2) = \Delta H(T_1) \]

### Step 4: Calculate the Enthalpy at T2
Thus, we have:
\[ \Delta H(373 K) = \Delta H(273 K) = -3.57 \text{ kJ} \]

### Conclusion
The enthalpy of the reaction at 373 K is also -3.57 kJ.

### Final Answer
\[ \Delta H(373 K) = -3.57 \text{ kJ} \]